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## Problem

Consider the family of functions

$x^p \ln(x)$.

Part 1: Calculate the indefinite integral

$\int x^p \ln(x) dx$.

Part 2: For what values $p$ does the following integral converge?

$\int_0^1 x^p \ln(x) dx$

## Solution

Part 1

This family of functions is the product of a power function with the natural logarithm function. Therefore the method of integration by parts should be used.

According to LIATE, let $u = \ln(x)$ and $dv = x^p dx$.

Then $du = \frac{1}{x}dx$ and $\frac{{x}^{p+1}}{p+1}$.

Therefore,

$\int x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p+1}}{p+1} \frac{1}{x} dx$
$\int x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p}}{p+1} dx$
$\int x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} - \frac{{x}^{p+1}}{(p+1)^2} + C$

Part 2

From the result of the indefinite integral, it is obvious that $p \ne -1$. Just look at the denominators. Now this means

$\int_0^1 x^p \ln(x) dx = \ln(x) \frac{{x}^{p+1}}{p+1} \Biggr|_0^1 - \frac{{x}^{p+1}}{(p+1)^2} \Biggr|_0^1$.

Notice that if $p < -1$, the power functions become reciprocal functions, which means the integral diverges when evaluating the lower bound. However if $p > -1$, the integral converges to the expression

$\int_0^1 x^p \ln(x) dx = \frac{-1}{(p+1)^2}$.

Therefore the integral

$\int_0^1 x^p \ln(x) dx$

converges if $p > -1$.

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